Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<climits>
#include<iostream>
#include<sstream>
#include<utility>
#include<map>
#include<vector>
#include<queue>
#include<algorithm>
#include<set>
#include<stack>

#define MAX 100009
using namespace std;
typedef pair<int,int> P;

typedef struct Node
{
	int n;
	vector<int>e;
}Node;

Node G[MAX];
int N,M,s,t,d,a,b;
long long res=0;
map<int,long long>SC,GC;
void dijcstra(int s,int g,map<int,long long>&X)
{
	int C[MAX];
	memset(C,0x7f,sizeof(C));
	P tmp(0,s);
	priority_queue<P,vector<P>,greater<P> >Q;
	Q.push(tmp);
	while(!Q.empty())
	{
		tmp=Q.top();Q.pop();
		if(tmp.first>=C[tmp.second])
			continue;
		C[tmp.second]=tmp.first;
		for(int i=0;i<G[tmp.second].e.size();i++)
		{
			int nn=G[tmp.second].e[i];
			if(C[nn]>tmp.first+1)
			{
				Q.push(make_pair(tmp.first+1,nn));
			}
		}
	}
	if(g==t)d=C[t];
	for(int i=0;i<N;i++)
		X[C[i]]++;
}

int main()
{
	scanf("%d%d%d%d",&N,&M,&s,&t);
	s--;t--;
	for(int i=0;i<M;i++)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		x--;y--;
		G[x].e.push_back(y);
		G[y].e.push_back(x);
	}
	dijcstra(s,t,SC);
	dijcstra(t,s,GC);
	for(int i=0;i<=d;i++)
	{
		if(d-i-2>=0)
			res+=SC[i]*GC[d-i-2];
	}
	printf("%lld\n",res);
	return 0;
}